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(C) Let the side length of the equilateral triangle $ABC$ be $a$. The distance between the parallel lines $L_1$ and $L_2$ is $6 + 3 = 9$ units.Let $

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$21 \sqrt{3}$

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(C) Let the side length of the equilateral triangle $ABC$ be $a$. The distance between the parallel lines $L_1$ and $L_2$ is $6 + 3 = 9$ units.Let $	heta$ be the angle that the side $BC$ makes with the line $L_2$. Since $C$ lies on $L_2$, the vertical distance from $A$ to $L_2$ is $9$. In the right-angled triangle formed by the projection of $A$ onto $L_2$, we have $\sin 	heta = \frac{3}{a}$ and $\sin(60^{\circ} + 	heta) = \frac{9}{a}$.Expanding $\sin(60^{\circ} + 	heta) = \sin 60^{\circ} \cos 	heta + \cos 60^{\circ} \sin 	heta = \frac{\sqrt{3}}{2} \cos 	heta + \frac{1}{2} \sin 	heta = \frac{9}{a}$.Substituting $\sin 	heta = \frac{3}{a}$ and $\cos 	heta = \sqrt{1 - \frac{9}{a^2}}$, we get $\frac{\sqrt{3}}{2} \sqrt{1 - \frac{9}{a^2}} + \frac{1}{2} \cdot \frac{3}{a} = \frac{9}{a}$.$\frac{\sqrt{3}}{2} \sqrt{\frac{a^2 - 9}{a^2}} = \frac{9}{a} - \frac{3}{2a} = \frac{15}{2a}$.$\sqrt{3} \sqrt{a^2 - 9} = 15 \implies 3(a^2 - 9) = 225 \implies a^2 - 9 = 75 \implies a^2 = 84$.The area of the equilateral triangle is $\frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} 	imes 84 = 21 \sqrt{3}$ sq. units.

Let a point $A$ lie between the parallel lines $L_1$ and $L_2$ such that its distances from $L_1$ and $L_2$ are $6$ and $3$ units, respectively. Then the area (in sq. units) of the equilateral triangle $ABC$, where the points $B$ and $C$ lie on the lines $L_1$ and $L_2$ respectively, is:

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